A lovely function problem

Algebra Level 5

A function $f:\mathbb{R} -\{a_1,a_2\} \to \mathbb{R},$ where $\mathbb{R}$ is the set of Real numbers, is defined by $f(x)=\frac{Ax^2+6x-8}{A+6x-8x^2}$ How many integral values of $A$ exist for which $f(x)$ is onto?

Details : $a_1,a_2$ are the roots of the quadratic equation $A+6x-8x^2=0$ .

The answer is 11.

2 solutions

U Z
Dec 19, 2014

$A = a$

$y = \dfrac{ax^{2} + 6x - 8}{a + 6x - 8x^2}$

$(a + 8y)x^2 + 6(1 - y)x - (8 + ay) = 0$

$36(1 - y)^{2} + 4(a - 8y)(8 + ay) \geq 0$ ( as x is real)

$9(1 - y)^{2} + (a - 8y)(8 + ay) \geq 0$

$= y^{2}( 9 + 8a) + y(46 + a^2 ) + ( 9 + 8a) \geq 0$ ( this is true for $y \in R , 9 + 8a>0 , ((46 + a^2 )^2 - 4( 9 + 8a)^2 \leq 0)$

Thus,

$a \geq \dfrac{-8}{9} , ( 46 + a^2 - 2(9 + 8a))(46 + a^2 + 2(9 + 8a)) \leq 0$

$= a \geq \dfrac{-8}{9} , ( a^2 - 16a + 28)(a^2 + 16a + 64) \leq 0$

$= a \geq \dfrac{-8}{9} , (a - 2)(a - 14)(a + 8)^2 \leq 0$

$= a \geq \dfrac{-8}{9} , 2\leq a \leq14$

All the solvers you too help

Close, but not quite. You need to be careful to justify each of the steps taken.

Main concern:
In order to justify the 3rd line, you still need to account for the fact that the denominator is non-zero. Otherwise any solution could be invalid. For example, with $a = 2, y = - \frac{1}{4}$ ,then the quadratic equation will be linear, and the solution comes at $x = 1$ . However, $\frac{ 2 + 6 - 8 } { 2 + 6 - 8 } \neq - \frac{1}{4}$ .
The third equation is true "because" we multiplied by 0, and so not all solutions to the third equation, are solutions to the second equation.

Note: typo in 2nd line, should be $8x^2$ .

I believe the answer should be 11.

Calvin Lin Staff - 6 years, 5 months ago

Ok , but here we are given that f(x) is a onto function and Sir how can you say that for a = 1 , y = -1/4? Thank you

U Z - 6 years, 5 months ago

I am saying that for $a = 2$ , then we do not have $f(x) = - \frac{1}{4}$ , and hence the function is not onto.

@Sandeep Bhardwaj Comments? I believe that $a = 2, 14$ do not work.

Calvin Lin Staff - 6 years, 5 months ago

@Calvin Lin – understood agree with you , 2 roots of the quadratic

U Z - 6 years, 5 months ago

@U Z – @megh choksi @Sandeep Bhardwaj For $a =2$ , what is the value of $x$ such that $f(x) = - \frac{1}{4}$ ?

Note that $f(1) = \frac{0}{0}$ . Even if you take limits of as x approaches 1, you do not get $- \frac{1}{4}$ . The reason why $x = 1$ is a solution, is because the denominator is 0, and when you multiply an equation (2nd line) by 0 (3rd line), then it becomes trivially true.

Calvin Lin Staff - 6 years, 5 months ago

@Calvin Lin – $f(x) = \dfrac{(x + 4)(x - 1)}{(x - 1)( - 4x - 1)}$

I did the same mistake earlier too in this problem

U Z - 6 years, 5 months ago

@Calvin Lin – we get -1/4 when we take limit as x approaches infinity.

Adit Mohan - 6 years, 5 months ago

@Calvin Lin – I apologize for late response, I was quite busy somewhere else.

I think there is no problem with taking $A=2,14$ . I am trying to explain it...

If we take $A=2$ , and lets assume $f(x)=y$ ,then $y=\dfrac{x^2+3x-4}{-4x^2+3x+1}$ .

$\implies (1+4y)x^2+(3-3y)x-4-y=0$

For $x$ to be real , its discriminant(say $D$ ) is to be $\geq 0$ .

$D=(3-3y)^2-4.(1+4y)(-4-y) \geq 0$

$\implies 25(y+1)^2 \geq 0$

$y \in \mathbb{R}$ .

Same happens when we take $A=14$ .

So Range=co-domain which tells us the function to be onto. @Calvin Lin @megh choksi

Sandeep Bhardwaj - 6 years, 5 months ago

@Sandeep Bhardwaj – And the above explanation is absolute. Before you say about why $y \neq \frac{-1}{4}$ , first explain why does the above explanation tells us so.. And if $A=2$ , $y$ will take the value $\frac{-1}{4}$ at $x=1$ .However, I know it will be $\frac{0}{0}$ on doing so. But if we cancel $(x-1)$ from Numerator and denominator both, then $y=\frac{-1}{4}$ . What you think about it ? @Calvin Lin

Sandeep Bhardwaj - 6 years, 5 months ago

@Sandeep Bhardwaj – Sir roughly the graph for a= 2 will be such kind

cancelling it we are making it a different function and

$Domain ~ of ~ \dfrac{f(x)}{g(x)} = Domain ~ f(x) \cap ~ Domain ~ g(x) , ~ g(x) \neq 0$ , So calvin lin Sir is correct

Sir is this explanation correct? @Sandeep Bhardwaj

U Z - 6 years, 5 months ago

@Calvin Lin – For A = 2, -0.25 < x < 1 while $-\infty$ < f (x) < -1;

For A = 14, -1 < x < 1.75 while -1 < f (x) < $\infty$ ;

Lu Chee Ket - 5 years, 6 months ago

Can you imagine I got this question in test today! Changed values, $f(x)=\dfrac{px^2+3x-4}{p+3x-4x^2}$

Only question I got wrong today! :-|

Pranjal Jain - 6 years, 4 months ago

I have updated the answer to 11. Can you please update the solution accordingly? Thanks!

Calvin Lin Staff - 6 years, 5 months ago

There is a thing that I want to say about this. If you take $A=2$ , then $f(x)=\dfrac{x^2+3x-4}{-4x^2+3x+1}$ , then $x=1$ is not included in the domain of $f(x)$ . But $f$ is given as $f:\mathbb{R} \to \mathbb{R}$ which means the Domain of $f$ is $\mathbb{R}$ , but $x=1$ is not included in the domain, which implies that $f$ is not a function. And same case happens with whatever you will take the value of $A$ from $2$ to $14$ . So the correct answer should be $0$ that there exist no such integral values of $A$ . What you think about it ? @Calvin Lin @megh choksi

Sandeep Bhardwaj - 6 years, 5 months ago

Just found an another integral solution for a

$= a \geq \dfrac{-8}{9} , (a - 2)(a - 14)(a + 8)^2 \leq 0$

also implies a = -8. and gives y=1

Thanks for pointing out mine reasoning its incorrect , so what the correct one?

U Z - 6 years, 5 months ago

@U Z – But in your solution you've mentioned that $a \geq \frac{-8}{9}$ , So taking intersection will exclude the value $a=-8$ . But I am saying that whatever the integral value you take from $[2,14]$ , $f(x)$ is not even a function due to the reason that domain of $f$ is given $\mathbb{R}$ . So if $f$ is not a function then how can we talk about its onto nature ? So why the answer should not be $0$ ? Also have a look at my above comment. @megh choksi

Sandeep Bhardwaj - 6 years, 5 months ago

@Sandeep Bhardwaj – So in this case, the "correct" interpretation of the question is that it is a function $f : \mathbb{R} \backslash \{a_1, a_2 \} \rightarrow \mathbb{R}$ , since we are clearly dealing with rational functions. By being less pedantic with the terminology, we can ease out such cases.

Alternatively, you could define $a_1, a_ 2 = \frac{ -6 \pm \sqrt{ (-6)^2 - 4 ( -8) A } } { 2 ( - 8 ) }$ , to get rid of these troublesome points.

Calvin Lin Staff - 6 years, 5 months ago

@Calvin Lin – This is what i said in starting that the domain of the function should be restricted

U Z - 6 years, 5 months ago

@Sandeep Bhardwaj – oh sorry by mistake yes it will be

$a \in [2 , 14] \cup \{-8\}$

$= a \in [ 2 , 14]$

U Z - 6 years, 5 months ago

Exactly. This is what I had concluded.

Sudeep Salgia - 6 years, 5 months ago

@Sudeep Salgia – how did to came to a conclusion that only 11 integral solutions as you are saying that there are 0

U Z - 6 years, 5 months ago

Isn't it Too much overrated ? No offence !

Karan Shekhawat - 6 years, 5 months ago

What is the explanation according to you for 2 and 14 as not the values for the given condition?

U Z - 6 years, 5 months ago

For A = 2, -0.25 < x < 1 while $-\infty$ < f (x) < -1;

For A = 14, -1 < x < 1.75 while -1 < f (x) < $\infty$ ;

Lu Chee Ket - 5 years, 6 months ago

@Sanjeet Raria @Deepanshu Gupta @Pratik Shastri @Mvs Saketh - can you provide a better solution to this- @Steven Zheng @Sandeep Bhardwaj @Pranjal Jain @brian charlesworth @Keshav Tiwari

U Z - 6 years, 5 months ago

@Karan Shekhawat i guess you are right.

Aarya Patil - 7 months, 2 weeks ago

Lu Chee Ket
Nov 24, 2015

Result of numerical analysis showed that A = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13} are only integers for individual range for $-\infty < f (x) < \infty$ :

3   132 0.914854216 -2.914854216    -0.343070331    1.093070331
4   164 0.850781059 -2.350781059    -0.425390530    1.175390530
5   196 0.8 -2  -0.5    1.25
6   228 0.758305739 -1.758305739    -0.568729304    1.318729304
7   260 0.723179678 -1.580322535    -0.632782219    1.382782219
8   292 0.693000468 -1.443000468    -0.693000468    1.443000468
9   324 0.666666667 -1.333333333    -0.75   1.5
10  356 0.643398113 -1.243398113    -0.804247642    1.554247642
11  388 0.622623437 -1.168077982    -0.856107225    1.606107225
12  420 0.603912564 -1.103912564    -0.905868846    1.655868846
13  452 0.586934293 -1.048472755    -0.953768227    1.703768227

With above figures, each and every one can be plotted using specific A for range of example (-0.343, 1.093) with A = 3 and etc, which satisfy the fact that $-\infty < f (x) < \infty$ . Other A's shall give other ranges.

Therefore, there are exactly 11 integer A which satisfy for wanted range for f (x).

Answer: $\boxed{11}$

I noticed that people tried to discuss for A = 2 and A = 14.

To numerical analysis, the answer is NO!

For A = 2, -0.25 < x < 1 while $-\infty$ < f (x) < -1;

For A = 14, -1 < x < 1.75 while -1 < f (x) < $\infty$ ;

Which are obviously NOT satisfied.

Lu Chee Ket - 5 years, 6 months ago

A lovely function problem

The answer is 11.

2 solutions

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